Problem: Ikta has an extraordinary feeling for undirected graphs. Ikta likes the undirected graph G and its two-point s, t pair (G, s, t) that has the greatest "beauty". The "beauty" of a pair (G, s, t) is the edge e = \\ {u, v \\} (u and v are two different points of G), and the shortest path from s to t in G. Is the number of things whose length is greater than the length of the shortest path from s to t in the undirected graph with g plus e.

Your job is to write a program that seeks its "beauty" given a pair (G, s, t).



Input

The input is given in the following format.

> N M s t
> x1 y1
> ...
> xi yi
> ...
> xM yM
>

First, the number of vertices, the number of edges, and the integers N, M, s, t representing the two vertices of the undirected graph are input. From the 2nd line to the M + 1th line, two vertices connected by edges are input. (However, let the set of G vertices be \\ {1, ..., N \\}.)

Constraints

Each variable being input satisfies the following constraints.

* 2 ≤ N ≤ 100,000
* 1 ≤ M ≤ 300,000
* 1 ≤ s, t, xi, yi ≤ N
* Different from s and t
* Guaranteed to reach t from s

Output

When the given graph is G, output the "beauty" of the set (G, s, t) in one line.

Examples

Input

3 2 1 3
1 2
2 3


Output

1


Input

9 8 7 8
2 6
4 9
8 6
9 2
3 8
1 8
8 5
7 9


Output

7


Input

4 3 1 4
1 2
3 4
4 1


Output

0


Input

9 7 8 9
9 6
6 5
3 6
3 7
2 5
8 5
1 4


Output

2

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
using vi=vector<int>;
using vvi=vector<vi>;
using vs=vector<string>;
using msi=map<string,int>;
using mii=map<int,int>;
using pii=pair<int,int>;
using vlai=valarray<int>;
using ll=long long;
#define rep(i,n) for(int i=0;i<n;i++)
#define range(i,s,n) for(int i=s;i<n;i++)
#define all(a) a.begin(),a.end()
#define rall(a) a.rbegin(),a.rend()
#define fs first
#define sc second
#define pb push_back
#define eb emplace_back
#define mp make_pair
#define INF 1e9
#define EPS 1e-9

int main(){
    int n,m,s,t;cin>>n>>m>>s>>t;
    vvi edge(n+1);
    rep(i,m){
        int a,b;cin>>a>>b;
        edge[a].pb(b);
        edge[b].pb(a);
    }

    vi depth(n+1,-1);
    depth[s]=0;
    queue<int> que; que.push(s);
    int ms=0;
    while(!que.empty()){
        int f=que.front();
        ms=depth[f];
        for(auto to:edge[f]){
            if(depth[to]==-1){
                depth[to]=depth[f]+1;
                que.push(to);
            }
        }
        que.pop();
    }
    int mt=0;
    vi depth2(n+1,-1);
    depth2[t]=0;
    que.push(t);
    while(!que.empty()){
        int f=que.front();
        mt=depth2[f];
        for(auto to:edge[f]){
            if(depth2[to]==-1){
                depth2[to]=depth2[f]+1;
                que.push(to);
            }
        }
        que.pop();
    }
    vvi ds(ms+1);
    range(i,1,n+1){if(depth[i]!=-1)ds[depth[i]].pb(i);}
    //for(auto l:ds){for(auto i:l)cout<<i<<",";cout<<endl;}
    vvi dt(mt+1);
    range(i,1,n+1){if(depth2[i]!=-1)dt[depth2[i]].pb(i);}
    //for(auto l:dt){for(auto i:l)cout<<i<<",";cout<<endl;}
    
    ll ans=0;
    for(int i=0;i<depth[t]-1;i++){
        ans+=ds[i].size()*dt[depth[t]-2-i].size();
    }

    cout<<ans<<endl;
    return 0;
}